Question

所以我得到了一些漂亮的代码集来运行并在我给它一个KittenID时找到我的信息，但它根本不起作用，我很伤心。哦，太累了，谁能告诉我哪里出错了？是的，我确实有：

<?php

date_default_timezone_set('America/New_York');
//If statements:

//find:
date_default_timezone_set('America/New_York');
if(isset($_POST['Find'])) 
{
            $connection = mysql_connect("ocelot.aul.fiu.edu","userName","password");

            // Check connection
            if (!$connection)
        {
            echo "Connection failed: " . mysql_connect_error();
        }
            else
            {  
                //select a database
                $dbName="spr15_xgotz001";
                $db_selected = mysql_select_db($dbName, $connection); 

                //confirm connection to database
                if (!$db_selected)
                {
                    die ('Can\'t use $dbName : ' . mysql_error());
                }
                else
                {
                    $result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID =<?php$_POST[KittenID]?>;)

                    while($row = mysql_fetch_array($result))
                    {
                    $Name = $row['Name'];
                    $KittenID = $row['KittenID'];
                    $KittenAge = $row['KittenAge'];
                    $Email = $row['Email'];
                    $Comments = $row['Comments'];
                    $Gender = $row['Gender'];
                    $Personality = $row['Personality'];
                    $Activity = $row['Activity']; 
                    echo $row['Comments'];

                    }
                }
            }

            mysql_close($connection);
}


?>

Answer 1

使用

$result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID = " .$_POST['KittenID']);

而不是

$result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID =<?php$_POST[KittenID]?>;)

注意：请将mysqli _ 用于未来的项目

Answer 2

您需要提供更多上下文。你如何设置$ _GET ['id'] ..它实际上是存储为$ _GET ['KittenID']（例如https://yoursite.com?view&KittenID=1）。如果是的话......

您可以设置变量并声明'KittenID'

$kittenid = $_POST['KittenID'];

$result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID = $kittenid");

我建议提供更多背景信息。你遇到了什么错误？你的参数是什么样的？

Answer 3

使用

$result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID = " .$_SERVER['KittenID']);

而不是

$result = mysql_query($connection,"SELECT * FROM Kittenz WHERE KittenID =<?php$_POST[KittenID]?>;)

试图在PHP中做SELECT，但我明白了

3 个答案: